Question

1. What is the molarity of 6.4g of methanol dissolved in 50 moles of water2. If the density of the above solution at 25C is 1.03kg/L what is the molarity methanol solution at 25°C

Answer 1

To calculate the molarity of the methanol solution, we need to first find the number of moles of methanol. The volume of the solution is not provided in the question, so we cannot calculate the molarity of the methanol solution at 25°C based on the given information.

Step-by-step explanation:

To calculate the molarity of the methanol solution, we need to first find the number of moles of methanol. The molarity (M) is defined as moles of solute divided by liters of solution. Given that we have 6.4g of methanol, we can convert it to moles by dividing by the molar mass of methanol, which is 32.04g/mol. So, the number of moles of methanol is:

6.4g / 32.04g/mol = 0.2 mol

The volume of the solution is not provided in the question, so we cannot calculate the molarity of the methanol solution at 25°C based on the given information. The density of the solution does not directly provide the volume of the solution, which is necessary for molarity calculations.

Answer 2

1. 0.22 mol/dm³

2. 0.23 mol/dm³

Step-by-step explanation:

Molarity: This can be defined as the concentration of a substance in mol per dm³

1.

M = n/v ...................... Equation 1

M = molarity of methanol, n = number of mole of methanol, v = volume of the solvent/water/solution.

n = g/m.m

Where g = mass of methanol, m.m = molar mass of methanol.

g = 6.4 g, m.m = 32 g /mole.

n = 6.4/32 = 0.2 mole

and ,

v = m/d

Where m = mass of water, d = density of water

m = mass of 50 moles of water = 18×50 = 900 g.

d = 1000 g/dm³.

v = 900/1000 = 0.9 dm³

Substituting the value of v and n into equation 1

M = 0.2/0.9

M = 0.22 mol/dm³

2.

M = n/v

v = volume of the solution

v = (mass of methanol + mass of water in the solution)/density of the solution

v = [(900+6.4)/1000]/1.03

v = 0.9064/1.03

v = 0.88 dm³

n = 0.2 mole

Therefore,

M = 0.2/0.88

M = 0.23 mol/dm³